A) \[30\,C\]
B) \[40\,C\]
C) \[60\,C\]
D) \[80\,C\]
Correct Answer: D
Solution :
At neutral point \[k\times \frac{20}{{{(20\times {{10}^{-2}})}^{2}}}=k\times \frac{Q}{{{(40\times {{10}^{-2}})}^{2}}}\] Þ Q = 80 CYou need to login to perform this action.
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